The square root of $i$ ?

Unlike Finn Hulse's solution, I am directly assuming that $\dfrac{a+i}{\sqrt{b}}=\sqrt{i}$ .

Squaring both sides, we get that $\dfrac{a^2-1+2ai}{b}=i$ .

Multiplying both sides by $b$ we get $a^2-1+2ai=bi$ .

Since the RHS has real part $0$ we must have $a^2-1=0$ . Since $a>0$ , we must have $a=1$ .

Therefore, $2i=bi$ , so $b=2$ .

Our answer is therefore $1+2=\boxed{3}$

We know that

$i=\cos \frac{\pi}{2}+i\sin \frac{\pi}{2}$

$\sqrt{i}=(\cos \frac{\pi}{2}+i\sin \frac{\pi}{2})^\frac{1}{2}$

By DeMoivre's Formula ,

$\sqrt{i}=\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}$

$\sqrt{i}=\frac{1}{\sqrt2}+i\frac{1}{\sqrt2}$

$\sqrt{i}=\frac{1+i}{\sqrt2}$

Hence, $a=1$ and $b=2$ , $a+b=3$ .

Let $m+ni$ be the form that our answer takes (ignoring the format provided). If $m+ni$ is squared, we can set it equal to $i$ . First let us look at the expansion in terms of $n$ and $m$ . This produces $m^{2}-n^{2}+2mni$ . Because the real part is equal to zero, and $m^{2}-n^{2}$ is that real part, it is obvious that $m=\pm n$ . Let us look at the imaginary portion. $2mni=i \Longrightarrow 2mn=1 \Longrightarrow mn=1/2$ . By substitution, $m^{2}=1/2$ and $m=n$ . Thus, the equation is $\frac{1+i}{\sqrt{2}}$ (after simplifying and combining fractions). Thus $a=1$ and $b=2$ . Adding $1+2=\boxed{3}$ . I hope you enjoyed the history lesson! :D

sqrt(i) = sqrt[(1 + i^2 + 2i)/2]. = (1+i)/sqrt(2). So a=1 and b=2. So a+b = 3.(Ans)...

$\dfrac{a+i}{\sqrt{b}}=\sqrt{i}$

$a+i=\sqrt{b}\sqrt{i}$

$(a+i)^2=(\sqrt{b}\sqrt{i})^2$

$a^2+2ai+i^2=bi$

$a^2+2ai-1=bi$

$a^2-1=0$

$a=1$ considering that $a$ is positive.

$b=2a=2(1)=2$

$a+b=1+2=\boxed{3}$ .